how many different digit number are there which have digit 1,2,3,4,5,6,7,8, such that digit 1 appears exactly once

its answer is 4c1 *7^3 how? why 4c1 is use here .?

• But its not mentioned anywhere in the question about construction of 4-digit number..also for number arrangement we do use permutation as far as i know...so explain
• 9 years agoHelpfull: Yes(17) No(1)
• Digit 1 can appear in any of the 4 positions of a 4 digit number => 4C1
  
  Further,as repetition is allowed,the othe three digits can have any of the remaining 7 numbers
  
  So total no of different 4 digit numbers possible = 4C1 * 7 * 7 *7 = 4C1 * 7^3
  
• 9 years agoHelpfull: Yes(15) No(12)
• insufficient data
• 9 years agoHelpfull: Yes(7) No(2)
• it might be 4digit number
• 9 years agoHelpfull: Yes(3) No(0)
• supposed that 4-digit number we have to form
  1 can appear only at one place out of 4 = 4c1
  now 3 places are remaining and repetition is allowed so 7*7*7
  final answer : 4c1*7*7*7
  
• 9 years agoHelpfull: Yes(2) No(1)
• Your question is incomplete...it should be
  How many 4 digit numbers re there which have digits 1,2,3,4,5,6,7,8 such that 1 appears exactly once and repitition is allowed?
  Ans=7*7*7*4c1(1 can be in unit,tens a,thousand and hundreds place)
• 9 years agoHelpfull: Yes(2) No(0)
• it is not given its 4 digit no
• 9 years agoHelpfull: Yes(1) No(0)
• VINAY SIPANI
  why "Further,as repetition is allowed,the othe three digits can have any of the remaining 7 numbers"???
  plz xplain
  
  
• 9 years agoHelpfull: Yes(0) No(0)